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شائع علم المثلثات >

sin^2(x)cos^2(x)=((1-cos^4(x)))/8

الحلّ

sin^{2} (x) cos^{2} (x) = \frac{( 1 - cos ^{4} ( x ) )}{8}

الحلّ

x = 1.18319 \dots + 2 πn, x = 2 π - 1.18319 \dots + 2 πn, x = 1.95839 \dots + 2 πn, x = - 1.95839 \dots + 2 πn, x = 2 πn, x = π + 2 πn

درجات

x = 67.79234 \dots^{\circ} + 36 0^{\circ} n, x = 292.20765 \dots^{\circ} + 36 0^{\circ} n, x = 112.20765 \dots^{\circ} + 36 0^{\circ} n, x = - 112.20765 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

sin^{2} (x) cos^{2} (x) = \frac{( 1 - cos ^{4} ( x ) )}{8}

من الطرفين

\frac{1 - cos ^{4} ( x )}{8}

اطرح

sin^{2} (x) cos^{2} (x) - \frac{1 - cos ^{4} ( x )}{8} = 0

sin^{2} (x) cos^{2} (x) - \frac{1 - cos ^{4} ( x )}{8}

بسّط:

\frac{8 sin ^{2} ( x ) cos ^{2} ( x ) - 1 + cos ^{4} ( x )}{8}

sin^{2} (x) cos^{2} (x) - \frac{1 - cos ^{4} ( x )}{8}

sin^{2} (x) cos^{2} (x) = \frac{sin ^{2} ( x ) cos ^{2} ( x ) 8}{8}

:حوّل الأعداد لكسور

= \frac{sin ^{2} ( x ) cos ^{2} ( x ) \cdot 8}{8} - \frac{1 - cos ^{4} ( x )}{8}

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{sin ^{2} ( x ) cos ^{2} ( x ) \cdot 8 - ( 1 - cos ^{4} ( x ) )}{8}

sin^{2} (x) cos^{2} (x) \cdot 8 - (1 - cos^{4} (x))

وسٌع:

sin^{2} (x) cos^{2} (x) \cdot 8 - 1 + cos^{4} (x)

sin^{2} (x) cos^{2} (x) \cdot 8 - (1 - cos^{4} (x))

= 8 sin^{2} (x) cos^{2} (x) - (1 - cos^{4} (x))

- (1 - cos^{4} (x)) : - 1 + cos^{4} (x)

- (1 - cos^{4} (x))

افتح أقواس

= - (1) - (- cos^{4} (x))

فعّل قوانين سالب-موجب

- (- a) = a, - (a) = - a

= - 1 + cos^{4} (x)

= sin^{2} (x) cos^{2} (x) \cdot 8 - 1 + cos^{4} (x)

= \frac{8 sin ^{2} ( x ) cos ^{2} ( x ) - 1 + cos ^{4} ( x )}{8}

\frac{8 sin ^{2} ( x ) cos ^{2} ( x ) - 1 + cos ^{4} ( x )}{8} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

8 sin^{2} (x) cos^{2} (x) - 1 + cos^{4} (x) = 0

Rewrite using trig identities

- 1 + cos^{4} (x) + 8 cos^{2} (x) sin^{2} (x)

cos^{2} (x) + sin^{2} (x) = 1

:فعّل نطريّة فيتاغوروس

sin^{2} (x) = 1 - cos^{2} (x)

= - 1 + cos^{4} (x) + 8 cos^{2} (x) (1 - cos^{2} (x))

- 1 + cos^{4} (x) + 8 cos^{2} (x) (1 - cos^{2} (x))

بسّط:

- 7 cos^{4} (x) + 8 cos^{2} (x) - 1

- 1 + cos^{4} (x) + 8 cos^{2} (x) (1 - cos^{2} (x))

8 cos^{2} (x) (1 - cos^{2} (x))

وسٌع:

8 cos^{2} (x) - 8 cos^{4} (x)

8 cos^{2} (x) (1 - cos^{2} (x))

a (b - c) = ab - a c

: افتح أقواس بالاستعانة بـ

a = 8 cos^{2} (x), b = 1, c = cos^{2} (x)

= 8 cos^{2} (x) \cdot 1 - 8 cos^{2} (x) cos^{2} (x)

= 8 \cdot 1 \cdot cos^{2} (x) - 8 cos^{2} (x) cos^{2} (x)

8 \cdot 1 \cdot cos^{2} (x) - 8 cos^{2} (x) cos^{2} (x)

بسّط:

8 cos^{2} (x) - 8 cos^{4} (x)

8 \cdot 1 \cdot cos^{2} (x) - 8 cos^{2} (x) cos^{2} (x)

8 \cdot 1 \cdot cos^{2} (x) = 8 cos^{2} (x)

8 \cdot 1 \cdot cos^{2} (x)

8 \cdot 1 = 8 :

اضرب الأعداد

= 8 cos^{2} (x)

8 cos^{2} (x) cos^{2} (x) = 8 cos^{4} (x)

8 cos^{2} (x) cos^{2} (x)

a^{b} \cdot a^{c} = a^{b + c}

:فعّل قانون القوى

cos^{2} (x) cos^{2} (x) = cos^{2 + 2} (x)

= 8 cos^{2 + 2} (x)

2 + 2 = 4 :

اجمع الأعداد

= 8 cos^{4} (x)

= 8 cos^{2} (x) - 8 cos^{4} (x)

= 8 cos^{2} (x) - 8 cos^{4} (x)

= - 1 + cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x)

- 1 + cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x)

بسّط:

- 7 cos^{4} (x) + 8 cos^{2} (x) - 1

- 1 + cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x)

جمّع التعابير المتشابهة

= cos^{4} (x) + 8 cos^{2} (x) - 8 cos^{4} (x) - 1

cos^{4} (x) - 8 cos^{4} (x) = - 7 cos^{4} (x) :

اجمع العناصر المتشابهة

= - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

= - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

= - 7 cos^{4} (x) + 8 cos^{2} (x) - 1

- 1 - 7 cos^{4} (x) + 8 cos^{2} (x) = 0

بالاستعانة بطريقة التعويض

- 1 - 7 cos^{4} (x) + 8 cos^{2} (x) = 0

cos (x) = u :

على افتراض أنّ

- 1 - 7 u^{4} + 8 u^{2} = 0

- 1 - 7 u^{4} + 8 u^{2} = 0 : u = \frac{1}{7}, u = - \frac{1}{7}, u = 1, u = - 1

- 1 - 7 u^{4} + 8 u^{2} = 0

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

اكتب بالصورة الاعتياديّة

- 7 u^{4} + 8 u^{2} - 1 = 0

v^{2} = u^{4}

وكذلك

v = u^{2}

اكتب المعادلة مجددًا، بحيث أنّ

- 7 v^{2} + 8 v - 1 = 0

- 7 v^{2} + 8 v - 1 = 0

حلّ:

v = \frac{1}{7}, v = 1

- 7 v^{2} + 8 v - 1 = 0

حلّ بالاستعانة بالصيغة التربيعيّة

- 7 v^{2} + 8 v - 1 = 0

الصيغة لحلّ المعادلة التربيعيّة

a = - 7, b = 8, c = - 1

لـ

v_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 7 ) ( - 1 )}{2 ( - 7 )}

v_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 7 ) ( - 1 )}{2 ( - 7 )}

8^{2} - 4 (- 7) (- 1) = 6

8^{2} - 4 (- 7) (- 1)

- (- a) = a

فعّل القانون

= 8^{2} - 4 \cdot 7 \cdot 1

4 \cdot 7 \cdot 1 = 28 :

اضرب الأعداد

= 8^{2} - 28

8^{2} = 64

= 64 - 28

64 - 28 = 36 :

اطرح الأعداد

= 36

36 = 6^{2} :

حلّل العدد لعوامله أوّليّة

= 6^{2}

n a^{n} = a

:فعْل قانون الجذور

6^{2} = 6

= 6

v_{1, 2} = \frac{- 8 \pm 6}{2 ( - 7 )}

Separate the solutions

v_{1} = \frac{- 8 + 6}{2 ( - 7 )}, v_{2} = \frac{- 8 - 6}{2 ( - 7 )}

v = \frac{- 8 + 6}{2 ( - 7 )} : \frac{1}{7}

\frac{- 8 + 6}{2 ( - 7 )}

(- a) = - a

:احذف الأقواس

= \frac{- 8 + 6}{- 2 \cdot 7}

- 8 + 6 = - 2 :

اطرح/اجمع الأعداد

= \frac{- 2}{- 2 \cdot 7}

2 \cdot 7 = 14 :

اضرب الأعداد

= \frac{- 2}{- 14}

\frac{- a}{- b} = \frac{a}{b}

: استخدم ميزات الكسور التالية

= \frac{2}{14}

2 :

إلغ العوامل المشتركة

= \frac{1}{7}

v = \frac{- 8 - 6}{2 ( - 7 )} : 1

\frac{- 8 - 6}{2 ( - 7 )}

(- a) = - a

:احذف الأقواس

= \frac{- 8 - 6}{- 2 \cdot 7}

- 8 - 6 = - 14 :

اطرح الأعداد

= \frac{- 14}{- 2 \cdot 7}

2 \cdot 7 = 14 :

اضرب الأعداد

= \frac{- 14}{- 14}

\frac{- a}{- b} = \frac{a}{b}

: استخدم ميزات الكسور التالية

= \frac{14}{14}

\frac{a}{a} = 1

فعّل القانون

= 1

حلول المعادلة التربيعيّة هي

v = \frac{1}{7}, v = 1

v = \frac{1}{7}, v = 1

Substitute back

v = u^{2},

solve for

u

u^{2} = \frac{1}{7}

حلّ:

u = \frac{1}{7}, u = - \frac{1}{7}

u^{2} = \frac{1}{7}

x = f (a), - f (a)

الحلول هي

x^{2} = f (a)

لـ

u = \frac{1}{7}, u = - \frac{1}{7}

u^{2} = 1

حلّ:

u = 1, u = - 1

u^{2} = 1

x = f (a), - f (a)

الحلول هي

x^{2} = f (a)

لـ

u = 1, u = - 1

1 = 1

1

1 = 1

فعّل القانون

= 1

- 1 = - 1

- 1

1 = 1

فعّل القانون

= - 1

u = 1, u = - 1

The solutions are

u = \frac{1}{7}, u = - \frac{1}{7}, u = 1, u = - 1

u = cos (x)

استبدل مجددًا

cos (x) = \frac{1}{7}, cos (x) = - \frac{1}{7}, cos (x) = 1, cos (x) = - 1

cos (x) = \frac{1}{7}, cos (x) = - \frac{1}{7}, cos (x) = 1, cos (x) = - 1

cos (x) = \frac{1}{7} : x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (x) = \frac{1}{7}

Apply trig inverse properties

cos (x) = \frac{1}{7}

cos (x) = \frac{1}{7} :

حلول عامّة لـ

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (x) = - \frac{1}{7} : x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

cos (x) = - \frac{1}{7}

Apply trig inverse properties

cos (x) = - \frac{1}{7}

cos (x) = - \frac{1}{7} :

حلول عامّة لـ

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1 :

حلول عامّة لـ

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

x = 0 + 2 πn

حلّ:

x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1 :

حلول عامّة لـ

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

وحّد الحلول

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn, x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn, x = 2 πn, x = π + 2 πn

أظهر الحلّ بالتمثيل العشريّ

x = 1.18319 \dots + 2 πn, x = 2 π - 1.18319 \dots + 2 πn, x = 1.95839 \dots + 2 πn, x = - 1.95839 \dots + 2 πn, x = 2 πn, x = π + 2 πn

رسم

أمثلة شائعة

sin^3(x)+sin(x)=2sin^{22}(x)

sin^{3} (x) + sin (x) = 2 sin^{22} (x)

(1+tan^2(x))/(1+sec(x))=sec(x)

\frac{1 + tan ^{2} ( x )}{1 + sec ( x )} = sec (x)

-sin^2(x)+2cos(x)-2=0

- sin^{2} (x) + 2 cos (x) - 2 = 0

sin(5x-1)= 4/5

sin (5 x - 1) = \frac{4}{5}

sin^2(x)= 1/36

sin^{2} (x) = \frac{1}{36}